\(\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
\end{array}\qquad\)
\(\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\
\end{array}\)
\(\bf \begin{array}{llll}
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}\)
so if you notice yours \(\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D
\end{array}\)
now.. normally the function \(\bf 3.2cos&\left( \frac{5}{3}\theta \right)\)
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis
now, with D = 6.1, that moves the midline up vertically that much
now.. the period, well, B = 5/3, normal period of cosine is \(2\pi\)
so, the new period will be \(\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}\)
notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)
